Practice problem
Find the derivative of h(x)=sinh(cosh−1x)h(x)=\sinh(\cosh^{-1}x)h(x)=sinh(cosh−1x)
h′(x)=1x2−1h'(x)=\frac{1}{\sqrt{x^2-1}}h′(x)=x2−11
h′(x)=coshxx2−1h'(x)=\frac{\cosh x}{\sqrt{x^2-1}}h′(x)=x2−1coshx
h′(x)=x+sinhxx2−1h'(x)=x+\frac{\sinh x}{\sqrt{x^2-1}}h′(x)=x+x2−1sinhx
h′(x)=xx2−1h'(x)=\frac{x}{\sqrt{x^2-1}}h′(x)=x2−1x
Type a concept, like “slope” or “photosynthesis”.
